\(\widehat{BDC}\) = 70 \(^{\circ}\)
=> \(\widehat{BDA}\) = 180 \(^{\circ}\) - \(\widehat{BDC}\) = 110 \(^{\circ}\)
Xét ΔDAB có DA = DB
=> ΔDAB cân tại D
=> \(\widehat{DAB}\) = \(\frac{180 ^{\circ} - \widehat{BDA}}{2}\) = \(\frac{70^{\circ} }{2}\)= 35 \(^{\circ}\)
Ta có : ΔABC cân tại B có \(\widehat{A}\) = 35 \(^{\circ}\)
=> \(\widehat{ABC}\) = 110 \(^{\circ}\)
\(\widehat{DBC}\) = \(\widehat{ABC}\) - \(\widehat{ABD}\) =\(\widehat{ABC}\) - \(\widehat{DAB}\)= 110 \(^{\circ}\) - 35 \(^{\circ}\) = 75\(^{\circ}\)
=> \(\widehat{DBC}\) = 75\(^{\circ}\)